Escape Speed

The formula for escape velocity is: where is the gravitational constant, is the mass of the planet, and is the radius of the planet. From the formula, it is clear that . - As the ratio increases, the escape velocity increases. Hence, Statement I is correct.

- However, depends on as seen from the formula, so escape velocity is not independent of the radius of the planet.

Hence, Statement II is incorrect. Thus, the correct answer is .

To find the escape velocity on the moon in terms of the given escape velocity on the planet , let's use the formula for escape velocity:

where:

• is the gravitational constant.
• is the mass of the celestial body.
• is the radius of the celestial body.

We are given:

• The mass of the moon is times the mass of the planet.
• The diameter of the moon is times the diameter of the planet, hence the radius of the moon is also times the radius of the planet.

If we denote the mass of the planet as and the radius of the planet as , then:

• Mass of the moon,
• Radius of the moon,

The escape velocity on the planet is given by:

For the moon, the escape velocity is:

Simplify the expression for the escape velocity on the moon:

Since , we have:

Thus, the escape velocity on the moon is . Therefore, the correct answer is:

To solve this problem, we need to determine the minimum speed required for an object to escape from the gravitational pull of the Earth when it is initially placed at a distance of above the Earth's surface. Here, is the Earth's radius, is the mass of the Earth, and is the gravitational constant.

The total energy of the object at the distance of from the Earth's surface (a total of from the Earth's center) is given by the sum of its gravitational potential energy and kinetic energy. For the object to not return to Earth, its velocity must be such that the total mechanical energy at any point is zero or greater.

The gravitational potential energy of the object at (from the center of the Earth) is: .

The condition for the object to escape is that the total mechanical energy should be :

Substituting the known values, we have: ,

Solving for the kinetic energy term:

Cancelling from both sides and solving for gives:

Therefore, the minimum speed required to project the object so that it does not return to Earth is: .

Hence, the correct answer is: .